A geometric progression series has first term 2 with common ratio as -1/2(x + 1). When x becomes 1/3, find the sum of all odd-numbered terms of the series. Answer given as 18/5 - not sure that it is correct! Thanks
$\displaystyle \sum_{k=0}^{\infty}\left(\frac{-(x+1)}{2}\right)^k$
$\displaystyle =1-\frac{x+1}{2}+\frac{(x+1)^2}{4}-\frac{(x+1)^3}{8}+...$
So the odd numbered terms are $\displaystyle 1+\frac{(x+1)^2}{4}+\frac{(x+1)^4}{16}+...$ which are the first, third, and 5th terms and so on
So this is just a geometric series with the ratio = $\displaystyle \frac{(x+1)^2}{4}$ so when $\displaystyle x=\frac{1}{3}$
The ratio is $\displaystyle \frac{(\frac{1}{3}+1)^2}{4}=\frac{\frac{16}{9}}{4} =\frac{4}{9}$
So the sum of a geometric series with ratio = $\displaystyle \frac{4}{9}$ is $\displaystyle \frac{1}{1-\frac{4}{9}}=\frac{9}{5}$
Sorry I didn't see that the first term was 2, being as this is the case the answer must be doubled and we get the correct answer
plug in x = 1/3, you find that r = -2/3
now, we want all odd terms, first note that these are the positive terms, secondly, note that the common ratio between consecutive odd terms is r^2. i hope you can see why. hence, the odd terms form a geometric series with first term a = 2 and common ratio r = 4/9. you should be able to do the rest. we get 18/5 as desired